Calculate the Length of a Curve: y=(e^x+e^-x)/2 from x=0 to x=ln4

• John O' Meara
In summary: No, I got as far as \int u^2\frac{1}{e^{x}-e^{-x}}du but I cannot express e^x-e^{-x} in terms of u.Then back up. It's much simpler than you are making it out to be. I'm trying to get you to notice that the expression inside of the (1/2) power in my last posting is a PERFECT SQUARE.Yes, But what property of a perfect square am I to notice?That sqrt(a^2)=a. Simplify the integral before you start substituting.\frac{1}{4}\int_0^{In4}((e^x + e^{

John O' Meara

Find the length of the curve y=(e^x+e^-x)/2 from x=0 to x=ln4.
$$\frac{dy}{dx}= \frac{1}{2} \frac{d}{dx}e^x + \frac{1}{2} \frac{d}{dx}e^{-x}= \frac{1}{2}(e^x-e^{-x}) \\$$.
Therefore $$(\frac{dy}{dx})^2=\frac{1}{4}(e^x-e^{-x})\\$$ Therefore $$1+(\frac{dy}{dx})^2=\frac{1}{4}(e^{2x} -2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}\\$$.
Therefore $$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$. Do I solve this by substitution by letting $$u^2 = e^{2x}-2e^xe^{-x}+e{-2x}+4$$? Or what way should I approach it? Using the said substitution I get the following integral: $$\int u^2 \frac{1}{e^{2x}-e^{-2x}}du$$

It will be a far simpler problem if you use the fact that $$\frac{e^x+e^{-x}}{2}=\cosh x$$. Have you encountered hyperbolic functions?

No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

John O' Meara said:
No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

Well, all you need to know for this question is that $$\frac{d}{dx}\cosh x=\sinh x , \hspace{1cm} \frac{d}{dx}\sinh x=\cosh x$$ and the identity $$\cosh^2x-\sinh^2x=1$$

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You said it would be a far simpler problem to use the fact that $$\frac{e^x+e^{-x}}{2}=coshx$$. Does that imply that it can still be done without the hyperbolic function method. If it does I would like to see the other longer way of solving this integral, please. Thanks.

$$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$

Start here. e^x*e^(-x)=1, right?

Yes, I got as far as $$\int u^2\frac{1}{e^{x}-e^{-x}}du$$ but I cannot express $$e^x-e^{-x}$$ in terms of u.

Then back up. It's much simpler than you are making it out to be. I'm trying to get you to notice that the expression inside of the (1/2) power in my last posting is a PERFECT SQUARE.

Yes, But what property of a perfect square am I to notice?

That sqrt(a^2)=a. Simplify the integral before you start substituting.

$$\frac{1}{4}\int_0^{In4}((e^x + e^{-x}})^2)^\frac{1}{2}dx$$, is the integral to be solved.

Yes. So you can eliminate the exponents and, voila, you don't need a substitution.

PS. You have an extra factor of (1/2) in there.

Thanks very much Dick.